Answer:
0.12234 m
0.40065 m/s
Explanation:
m = Mass of block = 1.4 kg
k = Spring constant = 16.5 N/m
x = Compresion of spring
v = Velocity of block
A = Amplitude
As the energy of the system is conserved we have
[tex]\frac{1}{2}mv^2=\frac{1}{2}kA^2\\\Rightarrow A=\sqrt{\frac{mv^2}{k}}\\\Rightarrow A=\sqrt{\frac{1.4\times 0.42^2}{16.5}}\\\Rightarrow A=0.12234\ m[/tex]
Amplitude of the oscillations is 0.12234 m
Again, as the energy of the system is conserved we have
[tex]\frac{1}{2}kA^2=\frac{1}{2}mv^2+\frac{1}{2}kx^2\\\Rightarrow v=\sqrt{\frac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\frac{16.5(0.12234^2-(0.3\times 0.12234)^2)}{1.4}}\\\Rightarrow v=0.40065\ m/s[/tex]
The block's speed at the point is 0.40065 m/s
Given values are:
The amplitude will be:
→ [tex]\frac{1}{2}mv^2 = \frac{1}{2} kA^2[/tex] (as the reserved energy of system)
or,
→ [tex]A = \sqrt{\frac{mv^2}{k} }[/tex]
By substituting the values,
[tex]= \sqrt{\frac{1.4\times 0.42^2}{16.5} }[/tex]
[tex]= 0.12234 \ m[/tex]
hence,
The speed of block will be:
→ [tex]\frac{1}{2}kA^2 = \frac{1}{2}mv^2 +\frac{1}{2} kx^2[/tex]
or,
→ [tex]v = \sqrt{\frac{k(A^2-x^2}{m} }[/tex]
[tex]= \sqrt{\frac{16.5(0.12234)^2-(0.3\times 0.12234)^2}{1.4} }[/tex]
[tex]= 0.40065 \ m/s[/tex]
Thus the above response is appropriate.
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